Price and greenness competition between two manufacturers with government intervention: a game-theoretic approach

Abstract

Recently, environmental concerns have drawn worldwide attention. As a result, the international communities have put pressure on both governments and industries to develop environmental conservation policies and green products, respectively. Also, the consumers’ environmental awareness has increased to the point that today’s consumers consider the greenness of the products in addition to the purchase price. In this study, we consider an industry consisting of two manufacturers that compete over selling price and carbon emissions, while the government levies a carbon emissions tax. We formulate the problem using a game-theoretic approach assuming the leader–follower relationship between the government and industry. The manufacturers maximize their profits, while the government maximizes three components: social welfare, environmental conservation, and revenues. We study the problem in two different settings: one in which the manufacturers have similar powers (Nash) and one in which one manufacturer is the leader in the market (Stackelberg). We find that superiority in production technology or a lower unit production cost does not necessarily lead to a lower selling price and that the outcome depends on the carbon emission tax rate. Based on a numerical experiment, government intervention reduces the total carbon emissions by more than almost half in both settings. We also found that an industry with Nash structure is preferred from the perspective of the consumers.

Appendices

Appendix

A Proof of Lemma 1

We need to show the joint concavity of the manufacturer decisions $p_i$ and $e_i$ on $\Pi _i^m(p_i,e_i)$. The first-order partial derivatives of $\Pi _i^m(p_i,e_i)$ are,

$$\begin{aligned} \frac{\partial \Pi _i^m(p_i,e_i)}{\partial p_i}= & {} (a_i-\beta _i p_i+\gamma p_{3-i}\\{} & {} -\delta _i e_i+\sigma e_{3-i} )-\beta _i(p_i-c_i-\tau e_i)\\= & {} a_i-2\beta _i p_i+\gamma p_{3-i}-(\delta _i-\tau \beta _i)e_i\\{} & {} +\sigma e_{3-i}+\beta _ic_i, \\ \frac{\partial \Pi _i^m(p_i,e_i)}{\partial e_i}= & {} -\tau (a_i-\beta _i p_i+\gamma p_{3-i}\\{} & {} -\delta _i e_i+\sigma e_{3-i} )-\delta _i(p_i-c_i-\tau e_i)+\eta _i(e_0-e_i )\\= & {} -\tau a_i+(\tau \beta _i-\delta _i)p_i-\tau \gamma p_{3-i}+(2\tau \delta _i-\eta _i)e_i\\{} & {} -\tau \sigma e_{3-i}+\delta _i c_i+\eta _i e_0, \end{aligned}$$

Then, four possible second-order derivatives of $\Pi _i^m(p_i,e_i)$ are, $\frac{\partial ^2 \Pi _i^m(p_i,e_i)}{\partial p_i \partial e_i}=\tau \beta _i-\delta _i,$ $\frac{\partial ^2 \Pi _i^m(p_i,e_i)}{\partial p_i^2}=-2\beta _i,$ $\frac{\partial ^2 \Pi _i^m(p_i,e_i)}{\partial e_i^2}=2\tau \delta _i-\eta _i,$ $\frac{\partial ^2 \Pi _i^m(p_i,e_i)}{\partial e_i p_i}=\tau \beta _i-\delta _i.$ As a result, the Hessian matrix of $\Pi ^m_i(p_i,e_i)$ is,

$$\begin{aligned} H(p_i,e_i)={} & {} \left[ \begin{array}{cc}{\frac{\partial ^{2} \Pi _{i}^{m}(p_i,e_i)}{\partial p_{i}^{2}}} &{} {\frac{\partial ^{2} \Pi _{i}^{m}(p_i,e_i)}{\partial p_{i} \partial e_{i}}} \\ {\frac{\partial ^{2} \Pi _{i}^{m}(p_i,e_i)}{\partial e_{i} \partial p_{i}}} &{} {\frac{\partial ^{2} \Pi _{i}^{m}(p_i,e_i)}{\partial e_{i}^{2}}}\end{array}\right] \\ ={} & {} \left[ \begin{array}{cc}{-2 \beta _{i}} &{} {\tau \beta _{i}-\delta _{i}} \\ {\tau \beta _{i}-\delta _{i}} &{} {2 \tau \delta _{i}-\eta _{i}}\end{array}\right] ,\quad i=1,2, \end{aligned}$$

and its determinant is

$$\begin{aligned} |H(p_i,e_i)|= & {} \left| \begin{array}{cc}{-2 \beta _{i}} &{} {\tau \beta _{i}-\delta _{i}} \\ {\tau \beta _{i}-\delta _{i}} &{} {2 \tau \delta _{i}-\eta _{i}}\end{array}\right| \\= & {} -2 \beta _{i}\left( 2 \tau \delta _{i}-\eta _{i}\right) -\left( \tau \beta _{i}-\delta _{i}\right) ^{2}\\= & {} 2 \beta _{i} \eta _{i}-\left( \tau \beta _{i}+\delta _{i}\right) ^{2}. \end{aligned}$$

Under the assumption $2 \beta _{i} \eta _{i}>\left( \tau \beta _{i}+\delta _{i}\right) ^{2}$, we have $|H(p_i,e_i)|>0$. Since the first entry of Hessian matrix is negative and its determinant is positive, Hessian matrix is negative-definite and $\Pi _i^m(p_i,e_i)$ is jointly concave in $p_i$ and $e_i$.

B Proof of Proposition 1

We check the general conditions proposed by Matsumoto and Szidarovszky (2016), under which a static game has at least one NE. Here are the conditions:

• 1. For each player, the strategy set is non-empty, convex and compact. For $i=1,2$, the strategy set of manufacturer i equals $[0,p_i^\textrm{max}]\times [0,e_0]$ which is non-empty, convex, and compact.

• 2. For each player, the payoff function is continuous in its variable domain. For $i=1,2$, the profit function $\Pi _i^m(p_i,e_i)$ is continuous in $p_i$ and $e_i$.

• 3. For each player, the payoff function is concave in terms of its decision variables assuming that the other players’ decision variables are given and fixed. Lemma 1 indicates that for $i=1,2$, $\Pi _i^m(p_i,e_i)$ is jointly concave in $p_i$ and $e_i$.

Therefore, all three conditions hold, and the static game between the manufacturers has at least one NE.

C Proof of Corollary 1

Lemma 1 shows the joint concavity of $\Pi _i^m (p_i,e_i)$, in $p_i$ and $e_i$. Therefore, the best response functions can be obtained from the first-order conditions (FOC). By putting the first-order partial derivatives of $\Pi _i^m(p_i,e_i)$ to zero, we have

$$\begin{aligned} \frac{\partial \Pi _i^m(p_i,e_i)}{\partial p_i}= & {} a_i-2\beta _i p_i+\gamma p_{3-i}-(\delta _i-\tau \beta _i)e_i\\{} & {} +\sigma e_{3-i}+\beta _ic_i=0 \\ \frac{\partial \Pi _i^m(p_i,e_i)}{\partial e_i}= & {} -\tau a_i+(\tau \beta _i-\delta _i)p_i-\tau \gamma p_{3-i}\\ {}{} & {} +(2\tau \delta _i-\eta _i)e_i -\tau \sigma e_{3-i}+\delta _i c_i+\eta _i e_0=0 \end{aligned}$$

The equations presented in Corollary 1 are the solution of $\left( \frac{\partial \Pi _i^m(p_i,e_i)}{\partial p_i},\frac{\partial \Pi _i^m(p_i,e_i)}{\partial e_i}\right) =(0,0)$.

D Proof of Proposition 2

NE is the intersection of the best response functions of all the players. Therefore, by solving the following system of equations consisting of the best response functions provided in Corollary 1, for $i=1,2$, NE solution could be obtained.

$$\begin{aligned} \left\{ \begin{array}{l}{p_{i}-\frac{\left( \sigma {{e}_{3-i}}+\gamma {{p}_{3-i}} \right) {{\xi }_{i}}+{{\omega }_{i}}}{{{\psi }_{i}}}=0,\,\,\,\,\,\,\,\,\,\,\,} \\ {{e}_{i}}-\frac{\left( \sigma {{e}_{3-i}}+\gamma {{p}_{3-i}} \right) {{\Omega }_{i}}+{{\nu }_{i}}}{{{\psi }_{i}}}=0,\,\,\,\,\,\,\,\,\,\,\end{array}\right. \end{aligned}$$

By doing some simple algebra, it could be shown that the coefficient matrix of the above system of equations is invertible. Therefore, its result is unique and it can be calculated as

$$\begin{aligned} p_{i}^{N}= \frac{\sigma \nu _{3-i} \xi _{i} \psi _{i}+\psi _{i} \psi _{3-i} \omega _{i}+\gamma \xi _{i} \psi _{i} \omega _{3-i}-\gamma \sigma \xi _{3-i} \omega _{i} \Omega _{i}-\sigma ^{2} \omega _{i} \Omega _{i} \Omega _{3-i}+\sigma \nu _{i} \xi _{i}\left( \gamma \xi _{3-i}+\sigma \Omega _{3-i}\right) }{\psi _{i}\left( \psi _{i} \psi _{3-i}-\left( \gamma \xi _{i}+\sigma \Omega _{i}\right) \left( \gamma \xi _{3-i}+\sigma \Omega _{3-i}\right) \right) }, \end{aligned}$$

and

$$\begin{aligned} e_{i}^{N}= \frac{\nu _{i}\left( \psi _{i} \psi _{3-i}-\gamma \xi _{i}\left( \gamma \xi _{3-i}+\sigma \Omega _{3-i}\right) \right) +\Omega _{i}\left( \sigma \nu _{3-i} \psi _{i}+\gamma \left( \psi _{i} \omega _{3-i}+\omega _{i}\left( \gamma \xi _{3-i}+\sigma \Omega _{3-i}\right) \right) \right) }{\psi _{i}\left( \psi _{i} \psi _{3-i}-\left( \gamma \xi _{i}+\sigma \Omega _{i}\right) \left( \gamma \xi _{i}+\sigma \Omega _{3-i}\right) \right) }. \end{aligned}$$

E Proof of Definition 1

Based on the definition of customer surplus (Phillips 2005), we have

$$\begin{aligned} CS=\sum _{i=1}^{2}\left( \int _{p_{i}^N}^{p_{i}^{maxN}} (a_i-\beta _i p_i+\gamma p_{3-i}^N-\delta _i e_i^N+\sigma e_{3-i}^N)dp_i\right) ,\nonumber \\ \end{aligned}$$

(8)

where $p_i^N$ is the equilibrium selling price of manufacturer i and is substituted from Proposition 3, $p_i^{maxN}$ is the saturation selling price of manufacturer i at equilibrium point that can be derived from Eq. (2) at equilibrium point. For $i=1,2$, using non-negativity property of demand $d_i\ge 0$ at equilibrium point, we have $a_i-\beta _i p_i^N+\gamma p_{3-i}^N-\delta _i e_i^N+\sigma e_{3-i}^N\ge 0.$ Then, $p_i^N\le \frac{a_i+\gamma p_{3-i}^N-\delta _i e_i^N+\sigma e_{3-i}^N}{\beta _i}.$ Hence, $p_i^{\textrm{max}N}=\frac{a_i+\gamma p_{3-i}^N-\delta _i e_i^N+\sigma e_{3-i}^N}{\beta _i},$ and $p_i^N\le p_i^{\textrm{max}N}$. As a result, Eq. (8) can be rewritten in terms of $p_i^{\textrm{max}N}$ as

$$\begin{aligned} CS=\sum _{i=1}^{2}\left( \beta _i\int _{p_{i}^N}^{p_{i}^{\textrm{max}N}} (p_i^{\textrm{max}N}- p_i)dp_i\right) , \end{aligned}$$

By doing some simplification, we get

$$\begin{aligned} \textrm{CS}=\frac{1}{2}\sum _{i=1}^{2}\beta _i\left( p_i^{\textrm{max}N}- p_i^{N}\right) ^2, \end{aligned}$$

F Proof of Proposition 3

The Hessian matrix of ${\overline{\Pi }}_2^m$ with respect to $p_2$ and $e_2$ equals $\left( \begin{array}{cc}{-2 \beta _{2}} &{} {\tau \beta _{2}-\delta _{2}} \\ {\tau \beta _{2}-\delta _{2}} &{} {2 \tau \delta _{2}-\eta _{2}}\end{array}\right)$, the first entry of which is negative. Furthermore,

$$\begin{aligned}{} & {} \left| \begin{array}{cc}{-2 \beta _{2}} &{} {\tau \beta _{2}-\delta _{2}} \\ {\tau \beta _{2}-\delta _{2}} &{} {2 \tau \delta _{2}-\eta _{2}}\end{array}\right| \\{} & {} \quad =-\left( \tau \beta _{2}+\delta _{2}\right) ^{2}+2 \beta _{2} \eta _{2} \ge 0 \Leftrightarrow 2 \beta _{2} \eta _{2} \ge \left( \tau \beta _{2}+\delta _{2}\right) ^{2} \end{aligned}$$

Therefore, the Hessian matrix is negative-definite which means that ${\overline{\Pi }}_2^m$ is jointly concave in $p_2$ and $e_2$. Hence, the optimal value of $p_2$ and $e_2$ is obtained by putting the gradient equal to zero.

G Proof of Proposition 4

To show the concavity of ${\overline{\Pi }}_1^m$, we should prove that its Hessian matrix is negative-definite. The Hessian matrix of ${\overline{\Pi }}_1^m$ is

$$\begin{aligned} \text{ Hessian }=\left( \begin{array}{cc}{-2 \beta _{1}+\frac{2 \gamma ^{2} \xi _{2}+2 \gamma \sigma \Omega _{2}}{\psi _{2}}} &{} {-\delta _{1}+\frac{\gamma \sigma _{2}^{*}}{\psi _{2}}+\frac{\sigma ^{2} \Omega _{2}}{\psi _{2}}-\tau \left( -\beta _{1}+\frac{\gamma ^{2} \xi _{2}}{\psi _{2}}+\frac{\gamma \sigma \Omega _{2}}{\psi _{2}}\right) } \\ {-\delta _{1}+\frac{\gamma \sigma \xi _{2}}{\psi _{2}}+\frac{\sigma ^{2} \Omega _{2}}{\psi _{2}}-\tau \left( -\beta _{1}+\frac{\gamma ^{2} \xi _{2}}{\psi _{2}}+\frac{\gamma \sigma \Omega _{2}}{\psi _{2}}\right) } &{} {-\eta _{1}-2 \tau \left( -\delta _{1}+\frac{\gamma \sigma \xi _{2}}{\psi _{2}}+\frac{\sigma ^{2} \Omega _{2}}{\psi _{2}}\right) }\end{array}\right) \end{aligned}$$

To be a negative-definite matrix, the first entry of the Hessian matrix should be negative and its determinant should be positive. The first entry is

$$\begin{aligned}{} & {} -2 \beta _{1}+\frac{2 \gamma ^{2} \xi _{2}+2 \gamma \sigma \Omega _{2}}{\psi _{2}}\nonumber \\{} & {} \quad =\frac{2 \Omega _{2}\left( \beta _{1} \Omega _{2}-\gamma (\sigma +\gamma \tau )\right) +2\left( \gamma ^{2}-2 \beta _{1} \beta _{2}\right) \eta _{2}}{2 \beta _{2} \eta _{2} -\left( \tau \beta _{2}+\delta _{2}\right) ^{2}} \le 0\nonumber \\ \end{aligned}$$

(9)

The latter inequality is correct from the assumption of the proposition. In addition, the determinant is

$$\begin{aligned}{} & {} \left( \frac{1}{\psi _{2}}\right) ^{2} \left( -2 \Phi _{2}\left( -\eta _{1} \psi _{2}-2 \tau \Phi _{3}\right) -\left( \Phi _{3}+\tau \Phi _{2}\right) ^{2}\right) \\{} & {} \quad =\left( \frac{1}{\psi _{2}}\right) ^{2}\left( 2 \eta _{1} \Phi _{2} \psi _{2}-\tau ^{2} \Phi _{2}^{2}+2 \tau \Phi _{2} \Phi _{3}-\Phi _{3}^{2}\right) \\{} & {} \quad = \left( \frac{1}{\psi _{2}}\right) ^{2}\left( 2 \eta _{1} \Phi _{2} \psi _{2}- \left( \Phi _{3}-\tau \Phi _{2}\right) ^{2} \right) \ge 0 \end{aligned}$$

Therefore, ${\overline{\Pi }}_1^m$ is jointly concave on $p_1$ and $e_1$. Hence, the optimal value of $p_1$ and $e_1$ could be obtained by solving the following system of equations.

$$\begin{aligned} \left\{ \begin{array}{l}{\frac{\partial {\overline{\Pi }}_{1}^{m}}{\partial p_{1}}=0} \\ {\frac{\partial {\overline{\Pi }}_{1}^{m}}{\partial e_{1}}=0}\end{array}\right. \end{aligned}$$

It is straightforward to obtain the solution of this equation system.

H Proof of Proposition 5

• 1. We need to show that if the condition of Proposition 5.1 holds, then $\frac{\partial p_i^N}{\partial \eta }<0$, otherwise $\frac{\partial p_i^N}{\partial \eta }\ge 0$. Since we are dealing with symmetric manufacturers, then for $i=1,2$, we have

  $$\begin{aligned} \frac{\partial p_{i}^{N}}{\partial \eta } =\frac{(\sigma +\beta \tau -\delta )(\delta +\beta \tau )\left( a+c(-\beta +\gamma )+(-\delta +\sigma -\beta \tau +\gamma \tau ) e_{0}\right) }{((-2 \beta +\gamma ) \eta +(\delta +\beta \tau )(\delta -\sigma +(\beta -\gamma ) \tau ))^{2}} \end{aligned}$$

  Since the denominator is positive, then $\frac{\partial p_i^N}{\partial \eta }\ge 0$ iff $\sigma +\beta \tau -\delta \le 0$ and $a+c(\gamma -\beta )+(-\delta +\sigma +(\gamma -\beta ) \tau ) e_{0} \le 0$. Hence, $\frac{\partial p_i^N}{\partial \eta }\ge 0$ iff $\tau \le \frac{\delta -\sigma }{\beta }$ and $\tau \le \frac{a}{e_{0}(\beta -\gamma )}-\frac{c}{e_{0}}-\frac{\delta -\sigma }{\beta -\gamma }$. As a result, if $\min (\frac{\delta -\sigma }{\beta },\frac{a}{e_{0}(\beta -\gamma )}-\frac{c}{e_{0}}-\frac{\delta -\sigma }{\beta -\gamma })\le \tau \le \max (\frac{\delta -\sigma }{\beta },\frac{a}{e_{0}(\beta -\gamma )}-\frac{c}{e_{0}}-\frac{\delta -\sigma }{\beta -\gamma })$, then $\frac{\partial p_i^N}{\partial \eta }\le 0$; otherwise, $\frac{\partial p_i^N}{\partial \eta }\ge 0$.

• 2. We need to show that if the condition of Proposition 5.2 holds, then $\frac{\partial e_i^N}{\partial \eta }<0$, otherwise $\frac{\partial e_i^N}{\partial \eta }\ge 0$. Since we are dealing with symmetric manufacturers, then for $i=1,2$, we have

  $$\begin{aligned} \frac{\partial e_{i}^{N}}{\partial \eta }=\frac{(2 \beta -\gamma )(\delta +\beta \tau )\left( a+c(-\beta +\gamma )+(-\delta +\sigma -\beta \tau +\gamma \tau ) e_{0}\right) }{((-2 \beta +\gamma ) \eta +(\delta +\beta \tau )(\delta -\sigma +(\beta -\gamma ) \tau ))^{2}} \end{aligned}$$

  Since the denominator is positive and $\beta \ge \gamma$, then $\frac{\partial e_i^N}{\partial \eta }\le 0$ iff $a+c(\gamma -\beta )+(-\delta +\sigma +(\gamma -\beta ) \tau ) e_{0}\le 0$. Hence, $\frac{\partial e_i^N}{\partial \eta }\le 0$ iff $\tau \le \frac{a}{e_{0}(\beta -\gamma )}-\frac{c}{e_{0}}-\frac{\delta -\sigma }{\beta -\gamma }$. As a result, if $\tau \le \frac{a}{e_{0}(\beta -\gamma )}-\frac{c}{e_{0}}-\frac{\delta -\sigma }{\beta -\gamma }$, then $\frac{\partial e_i^N}{\partial \eta }\le 0$; otherwise, $\frac{\partial e_i^N}{\partial \eta }\ge 0$.

• 3. We need to show that if the condition of Proposition 5.3 holds, then $\frac{\partial CS(p_1^N,p_2^N)}{\partial \eta }<0$; otherwise, $\frac{\partial CS(p_1^N,p_2^N)}{\partial \eta }\ge 0$. For $i=1,2$, we have

  $$\begin{aligned} \frac{\partial CS(p_1^N,p_2^N)}{\partial \eta }=\frac{2 \beta \eta (\delta +\beta \tau )(\delta -\sigma +(\beta -\gamma ) \tau )\left( -a+c(\beta -\gamma )+(\delta -\sigma +\beta \tau -\gamma \tau ) e_{0}\right) ^{2}}{((-2 \beta +\gamma ) \eta +(\delta +\beta \tau )(\delta -\sigma +(\beta -\gamma ) \tau ))^{3}} \end{aligned}$$

  Since $\beta \ge \gamma$ and $\delta \ge \sigma$, the numerator is non-negative. Then, $\frac{\partial CS(p_1^N,p_2^N)}{\partial \eta }<0$, if the denominator is negative. Hence,

  $$\begin{aligned}{} & {} (-2 \beta +\gamma ) \eta +(\delta +\beta \tau )(\delta -\sigma +(\beta -\gamma ) \tau )<0 \Leftrightarrow \\{} & {} \quad \tau <\frac{\delta (\beta -\gamma )+\beta (\delta -\sigma )-\sqrt{(\delta (\beta -\gamma )+\beta (\delta -\sigma ))^{2}-4 \beta (\beta -\gamma )(\delta (\delta -\sigma )+\eta (\gamma -2 \beta ))}}{2 \beta (\gamma -\beta )}={\overline{\tau }}. \end{aligned}$$

  As a result, if $\tau <{\overline{\tau }}$, then $\frac{\partial CS(p_1^N,p_2^N)}{\partial \eta }<0$; otherwise, $\frac{\partial CS(p_1^N,p_2^N)}{\partial \eta }\ge 0$.

I Proof of Proposition 6

• 1. We need to show that if the condition of Proposition 6.1 holds, then $\frac{\partial p_i^N}{\partial c}\le 0$, otherwise $\frac{\partial p_i^N}{\partial c}\ge 0$. Since we are dealing with symmetric manufacturers, then for $i=1,2$, we have

  $$\begin{aligned} \frac{\partial p_{i}^{N}}{\partial c}=\frac{-\beta \eta +(\delta -\sigma )(\delta +\beta \tau )}{(\gamma -2 \beta ) \eta +(\delta +\beta \tau )(\delta -\sigma +(\beta -\gamma ) \tau )}. \end{aligned}$$

  For the numerator, iff $\tau \le \frac{\eta }{\delta -\sigma }-\frac{\delta }{\beta }$, then $-\beta \eta +(\delta -\sigma )(\delta +\beta \tau )\ge 0$; otherwise, $-\beta \eta +(\delta -\sigma )(\delta +\beta \tau )\le 0$. For the denominator, iff

  $$\begin{aligned} \tau >\frac{\delta (\beta -\gamma )+\beta (\delta -\sigma )-\sqrt{(\delta (\beta -\gamma )+\beta (\delta -\sigma ))^{2}-4 \beta (\beta -\gamma )(\delta (\delta -\sigma )+\eta (\gamma -2 \beta ))}}{2 \beta (\gamma -\beta )}={\overline{\tau }}, \end{aligned}$$

  then $(\gamma -2 \beta ) \eta +(\delta +\beta \tau )(\delta -\sigma +(\beta -\gamma ) \tau )>0$; otherwise, $(\gamma -2 \beta ) \eta +(\delta +\beta \tau ) (\delta -\sigma +(\beta -\gamma ) \tau )<0$. Then, $\frac{\partial p_i^N}{\partial c}<0$ iff the numerator and denominator have different signs, i.e., either positive and negative or negative and positive. Hence, if $\min \left( \frac{\eta }{\delta -\sigma }-\frac{\delta }{\beta },{\overline{\tau }}\right) \le \tau \le \max \left( \frac{\eta }{\delta -\sigma }-\frac{\delta }{\beta },{\overline{\tau }}\right)$ then $\frac{\partial p_{i}^{N}}{\partial c}<0$; otherwise, $\frac{\partial p_i^N}{\partial c}>0$.

• 2. We need to show that if the condition of Proposition 6.2 holds, then $\frac{\partial e_i^N}{\partial c}\le 0$, otherwise $\frac{\partial e_i^N}{\partial c}\ge 0$. Since we are dealing with symmetric manufacturers, then for $i=1,2$, we have

  $$\begin{aligned} \frac{\partial e_{i}^N}{\partial c}=\frac{(\gamma -\beta )(\delta +\beta \tau )}{(\gamma -2 \beta ) \eta +(\delta +\beta \tau )(\delta -\sigma +(\beta -\gamma ) \tau )} \end{aligned}$$

  For the numerator, since $\beta \ge \gamma$, then for all $\tau >0$, we have $(\gamma -\beta )(\delta +\beta \tau )\le 0$. For the denominator, similar to part 1, iff $\tau >{\overline{\tau }}$, then $(\gamma -2 \beta ) \eta +(\delta +\beta \tau )(\delta -\sigma +(\beta -\gamma ) \tau )>0$; otherwise, $(\gamma -2 \beta ) \eta +(\delta +\beta \tau )(\delta -\sigma +(\beta -\gamma ) \tau )<0$. Then, iff $\tau >{\overline{\tau }}$ then $\frac{\partial e_{i}^N}{\partial c}<0$; otherwise, $\frac{\partial e_{i}^N}{\partial c}>0$.

• 3. We need to show that if the condition of Proposition 6.3 holds, then $\frac{\partial CS}{\partial c}\le 0$; otherwise, $\frac{\partial CS}{\partial c}\ge 0$. Since we are dealing with symmetric manufacturers, then for $i=1,2$ we have

  $$\begin{aligned} \frac{\partial C S}{\partial c}=\frac{2 \beta \eta ^{2}(\gamma -\beta ) \left( a+c(\gamma -\beta )+e_{0}(\sigma -\delta -(\beta -\gamma )\tau ) \right) }{\left( (\gamma -2 \beta ) \eta +(\delta +\beta \tau )(\delta -\sigma +(\beta -\gamma ) \tau )\right) ^{2}} \end{aligned}$$

  For all $\tau >0$, the denominator is positive. Since $\beta \ge \gamma$, iff $\tau \le \frac{a+c(\gamma -\beta )+e_0(\sigma -\delta )}{e_0(\beta -\gamma )}$, then $\frac{\partial CS}{\partial c}\le 0$; otherwise, $\frac{\partial C S}{\partial c}>0$.

J Proof of Proposition 7

We show that if the condition holds, then $\frac{\Pi _{1}^{m}({\varvec{p}}^N,{\varvec{e}}^N)}{{\overline{\Pi }}_{1}^{m}({\varvec{p}}^S,{\varvec{e}}^S)}<1$; otherwise $\frac{\Pi _{1}^{m}({\varvec{p}}^N,{\varvec{e}}^N)}{{\overline{\Pi }}_{1}^{m}({\varvec{p}}^S,{\varvec{e}}^S)}>1$. Based on Equations (3) and (6), we can write

$$\begin{aligned} \frac{\Pi _{1}^{m}({\varvec{p}}^N,{\varvec{e}}^N)}{{\overline{\Pi }}_{1}^{m}({\varvec{p}}^S,{\varvec{e}}^S)} =1-\frac{\eta (-\gamma \eta +\sigma (\delta +\beta \tau )+\gamma \tau (\delta +\beta \tau ))^{2}\left( 2 \beta \sigma ^{2}-2 \gamma \sigma (\delta -\beta \tau )+\gamma ^{2}(\eta -2 \delta \tau )\right) }{((-2 \beta +\gamma ) \eta +(\delta +\beta \tau )(\delta -\sigma +(\beta -\gamma ) \tau ))^{2}((2 \beta +\gamma ) \eta -(\delta +\beta \tau )(\delta +\sigma +(\beta +\gamma ) \tau ))^{2}} \end{aligned}$$

To have $\frac{\Pi _{1}^{m}({\varvec{p}}^N,{\varvec{e}}^N)}{{\overline{\Pi }}_{1}^{m}({\varvec{p}}^S,{\varvec{e}}^S)}<1$, both numerator and denominator should be positive. Since the denominator is positive, then $\frac{\Pi _{1}^{m}({\varvec{p}}^N,{\varvec{e}}^N)}{{\overline{\Pi }}_{1}^{m}({\varvec{p}}^S,{\varvec{e}}^S)}<1$ iff $(2 \beta \sigma ^{2}-2 \gamma \sigma (\delta -\beta \tau )+\gamma ^{2}(\eta -2 \delta \tau ))>0$. By doing some simple algebra, $\Pi _{1}^{m}({\varvec{p}}^N,{\varvec{e}}^N)<{{\overline{\Pi }}_{1}^{m}({\varvec{p}}^S,{\varvec{e}}^S)}$ iff $\gamma ^{2}\eta >2(\gamma \delta -\beta \sigma )(\gamma \tau +\sigma )$. Therefore, if $\gamma ^{2}\eta >2(\gamma \delta -\beta \sigma )(\gamma \tau +\sigma )$ holds, then $\Pi _{1}^{m}({\varvec{p}}^N,{\varvec{e}}^N)<{{\overline{\Pi }}_{1}^{m}({\varvec{p}}^S,{\varvec{e}}^S)}$; otherwise, $\Pi _{1}^{m}({\varvec{p}}^N,{\varvec{e}}^N)>{{\overline{\Pi }}_{1}^{m}({\varvec{p}}^S,{\varvec{e}}^S)}$.

K Proof of Proposition 8

• 1. We show that if the condition of Proposition 8.1 does not hold, then $p_{2}^{N}-p_{1}^{N}\ge 0$. From Corollary 1, we can write

  $$\begin{aligned} p_{2}^{N}-p_{1}^{N}=\frac{(-\beta \eta +(\delta +\sigma )(\delta +\beta \tau ))\left( c_{2}-c_{1}\right) }{-(2 \beta +\gamma ) \eta +(\delta +\beta \tau )(\delta +\sigma +(\beta +\gamma ) \tau )} \end{aligned}$$

  To have $p_{2}^{N}-p_{1}^{N}\ge 0$, then both numerator and denominator should be either positive or negative. Without loss of generality, we assume that both are negative. Since $c_2>c_1$, then $p_{2}^{N}-p_{1}^{N}\ge 0$ iff $\eta \ge \frac{(\delta +\sigma )(\delta +\beta \tau )}{\beta }$ and $\eta \ge \frac{(\delta +\beta \tau )(\delta +\sigma +(\beta +\gamma ) \tau )}{2 \beta +\gamma }$. Let define ${\underline{\eta }}=\frac{(\delta +\sigma )(\delta +\beta \tau )}{\beta }$ and ${\overline{\eta }}=\frac{(\delta +\beta \tau )(\delta +\sigma +(\beta +\gamma ) \tau )}{2 \beta +\gamma }$. As a result, if $\min ({\underline{\eta }},{\overline{\eta }})<\eta <\max ({\underline{\eta }},{\overline{\eta }})$, then $p_{2}^{N}-p_{1}^{N}<0$. Hence, if $\min ({\underline{\eta }},{\overline{\eta }})<\eta <\max ({\underline{\eta }},{\overline{\eta }})$, then $p_{2}^{N}< p_{1}^{N}$; otherwise, $p_{2}^{N}\ge p_{1}^{N}$.

• 2. We show that if the condition of Proposition 8.2 holds, then $e_{2}^{N}-e_{1}^{N}<0$. From Corollary 1, we can write

  $$\begin{aligned} e_{2}^{N}-e_{1}^{N}=-\frac{(\beta +\gamma )(\delta +\beta \tau )\left( c_{2}-c_{1}\right) }{-(2 \beta +\gamma ) \eta +(\delta +\beta \tau )(\delta +\sigma +(\beta +\gamma ) \tau )} \end{aligned}$$

  Since $c_2>c_1$, then the nominator is positive. To have $e_{2}^{N}-e_{1}^{N}<0$, then the denominator should be positive, too. Hence, $e_{2}^{N}-e_{1}^{N}<0$ iff $\eta \le \frac{(\delta +\beta \tau )(\delta +\sigma +(\beta +\gamma ) \tau )}{(2 \beta +\gamma )}$. As a result, if $\eta \le \frac{(\delta +\beta \tau )(\delta +\sigma +(\beta +\gamma ) \tau )}{(2 \beta +\gamma )}$, then $e_{2}^{N}<e_{1}^{N}$; otherwise, $e_{2}^{N}\ge e_{1}^{N}$.